Question

1. given the following information:sno₂ (s) + 2 h₂ (g) ↔ sn (s) + 2 h₂o (g) kc = 8.12h₂ (g) + co₂ (g) ↔ h₂o (g) + co (g) kc = 0.771calculate kc for the reaction: sno₂ (s) + 2 co (g) ↔ sn (s) + 2 co₂ (g)

Explanation:

Step1: Reverse the second reaction

Reverse $\ce{H2 (g) + CO2 (g) <=> H2O (g) + CO (g)}$ to get $\ce{H2O (g) + CO (g) <=> H2 (g) + CO2 (g)}$. When reversing a reaction, the new equilibrium constant is the reciprocal of the original:
$K_{c2}' = \frac{1}{0.771} \approx 1.297$

Step2: Multiply reversed reaction by 2

Multiply $\ce{H2O (g) + CO (g) <=> H2 (g) + CO2 (g)}$ by 2 to get $\ce{2H2O (g) + 2CO (g) <=> 2H2 (g) + 2CO2 (g)}$. When multiplying a reaction by a factor $n$, raise the equilibrium constant to the power $n$:
$K_{c2}'' = (1.297)^2 \approx 1.682$

Step3: Add first and modified reactions

Add $\ce{SnO2 (s) + 2 H2 (g) <=> Sn (s) + 2 H2O (g)}$ and $\ce{2H2O (g) + 2CO (g) <=> 2H2 (g) + 2CO2 (g)}$. When adding reactions, multiply their equilibrium constants:
$K_{c(total)} = K_{c1} \times K_{c2}''$
$K_{c(total)} = 8.12 \times 1.682$

Step4: Calculate final Kc

Compute the product:
$K_{c(total)} = 8.12 \times 1.682 \approx 13.66$

Answer:

$\approx 13.7$